Exponentials in $\operatorname{Fun}(\mathcal C^{\mathrm{op}}, \operatorname{Cat}_{\infty})$ Let $\mathcal C$ be a locally small $\infty$-category. In several places I encountered the claim that the functor $\infty$- category $\operatorname{Fun}(\mathcal C^{\mathrm{op}}, \operatorname{Cat}_{\infty})$
is Cartesian closed, with exponentials given by
$$G^F\colon c \mapsto \operatorname{Nat}(h_c\times F, G)$$
where $\operatorname{Nat}(-, -)$ denotes the $\infty$-category of natural transformations in $\operatorname{Fun}(\mathcal C^{\mathrm{op}}, \operatorname{Cat}_{\infty})$ (i.e. an enhancement of the mapping space in that category, a formal definition can be found in the paper of Gepner, Haugseng and Nikolaus) and where $h$ is the Yoneda embedding. I never found a proof of that result, though. Is anybody familiar with a reference for this claim?
 A: For the sake of not leaving this question unanswered: the $\infty$-category $\operatorname{Cat}_{\infty}$ arises as a localisation of the presheaf $\infty$-category $\operatorname{Fun}(\Delta^{\mathrm{op}}, \mathcal S)$ in which the localisation functor commutes with finite products. By postcomposition, we thus find that $\operatorname{Fun}(\mathcal C^{\mathrm{op}},\operatorname{Cat}_\infty)$ arises as a localisation of $\operatorname{Fun}((\mathcal C\times\Delta)^{\mathrm{op}},\mathcal S)$ where the localisation functor preserves finite products. Since presheaf $\infty$-categories are cartesian closed, this means that $\operatorname{Fun}(\mathcal C^{\mathrm{op}},\operatorname{Cat}_\infty)$ is an exponential ideal in $\operatorname{Fun}((\mathcal C\times\Delta)^{\mathrm{op}},\mathcal S)$ and therefore in particular cartesian closed. Moreover, since this result implies that the exponential in $\operatorname{Fun}(\mathcal C^{\mathrm{op}},\operatorname{Cat}_\infty)$ is computed in $\operatorname{Fun}((\mathcal C\times\Delta)^{\mathrm{op}},\mathcal S)$, the desired formula follows from the well-known fact that the exponential in presheaf $\infty$-categories is computed according to the very same formula.
